Sample chapter from GCSE Mathematics: Revision and Practice New 2010 Edition Higher Student Book, by...

5 Algebra 2

In this unit you will:

l learn how to change the subject of a formula

l learn how to solve inequalities and represent them as

regions on a graph

l solve problems in direct and indirect proportion

l identify and plot curved graphs

l use graphs to solve equations.

Functional skills coverage and range:

l Understand, use, and calculate ratio and proportion,

including problems involving scale

l Understand and use simple equations and simple formulae

involving one- or two-step operations.

5.1 Changing the subject of a formula

The operations that you use in solving ordinary linear equations are

exactly the same as the operations you need to change the subject

of a formula.

5.1.1 Simple formulaeE X AM P L E

Make x the subject in these formulae.

a ax� p = t b y(x + y) = v2

...........................................................................a ax� p = t b y(x + y) = v2

ax = t + p

x =t + pa

yx + y2 = v2

yx = v2 � y2

x =v2 � y2

y

Exercise 1��MMake x the subject.

1 x + b = e 2 x� t = m 3 x� f = a + b

4 x + h = A + B 5 x + t = y + t 6 a + x = b

7 k + x = m 8 v + x = w + y 9 ax = b

LinksIf you have an idea

of the shape of a

graph then you

can model a

situation to find

out the exact

equation, for

example, how fast

a radioactive

isotope is

decaying. This

could tell you

exactly what the

isotope is.

10 hx = m 11 mx = a + b 12 kx = c� d

13 vx = e + n 14 3x = y + z 15 xp = r

16 xm = h�m 17 ax + t = a 18 mx� e = k

19 ux� h = m 20 ex + q = t 21 kx� u2 = v2

Now do these.

22 gx + t2 = s2 23 xa + k = m2 24 xm� v = m

25 a + bx = c 26 t + sx = y 27 y + cx = z

28 a + hx = 2a 29 mx� b = b 30 kx + ab = cd

31 a(x� b) = c 32 c(x� d) = e 33 m(x + m) = n2

34 k(x� a) = t 35 h(x� h) = k 36 m(x + b) = n

37 a(x� a) = a2 38 c(a + x) = d 39 m(b + x) = e

In questions 31–39multiply out thebrackets first.

5.1.2 Formulae involving fractionsE X AM P L E


axa

= p bmx

= t ca2

m=

dx

Notice in parts band c that when xis on the bottomyou start bymultiplying bothsides by x.

...........................................................................

axa

= p bmx

= t ca2

m=

dx

x = ap m = xt xa2= dmmt= x x =

dma2


1 xt= m 2 x

e= n 3 x

p= a

4 am =xt

5 bc =xa

6 e =xy2

7 xa

= (b + c) 8 xt= (c� d) 9 x

m= s + t

10 xk

= h + i 11 xb

=ac

12 xm

=zy

13 xh

=cd

14 mn

=xe

15 be=

xh

16 x(a + b)

= c 17 x(h + k)

= m 18 xu

=my

19 x(h� k)

= t 20 x(a + b)

= (z + t) 21 t =ex

22 a =ex

23 m =hx

24 ab

=cx

When x is on thebottom, multiplyboth sides by x.

248 Algebra 2

25 ux

=cd

26 mx

= t 2 27 hx

= sin 20�

28 ex

= cos 40� 29 mx

= tan 46� 30 a2

b2=

c2

x

You will meetsine, cos and tanin chapter 6.

5.1.3 Formulae with negative x-termsE X AM P L E

Make x the subject of these formulae.

Notice that ineach question thefirst step is tomake the x-termpositive by takingit to the otherside.

a t � x = a2 b h� bx = m c a(m� x) = h...........................................................................a t � x = a2 b h� bx = m c a(m� x) = h

t = a2 + x h = m + bx am� ax = h

t � a2 = x h�m = bx am = h + ax

or x = t � a2 h�mb

= x am� h = ax

am� ha

= x


1 a� x = y 2 h� x = m 3 z� x = q 4 v = b� x

5 m = k� x 6 h� cx = d 7 y�mx = c 8 k� ex = h

9 a2 � bx = d 10 m2 � tx = n2 11 v2 � ax = w 12 y� x = y2

13 k� t 2x = m 14 e = b� cx 15 z = h� gx 16 a + b = c� dx

17 y2 = v2 � kx 18 h = d � fx 19 a(b� x) = c 20 h(m� x) = n

Make a the subject.

21 m(c 2 a) = t 22 v(p 2 a) = w 23 e = d (q 2 a) 24 b2 2 a = r2

25 x� af

= 2f 26 B� AaD

= E 27 D� EaN

= B 28h� fa

b= x

29 v2 � haC

= d 30M(a + B)

N= T 31

f (Na� e)

m= B 32

T(M � a)

E= F

Make x the subject (more difficult).

33 2x

+ 1 = 3y 34 5x� 2 = 4z 35 A

x+ B = C 36 V

x+ G = H

37 rx� t = n 38 q =

bx

+ d 39 t =mx� n 40 h = d � b

x

41 C � dx

= e 42 r � mx

= e2 43 t 2 = b� nx

44 dx

+ b = mn

45 3M = M +N

P + x46 A =

Bc + x

25A 47 m2

x� n = 2p 48 t = w� q

x

Changing the subject of a formula 249

5.1.4 Formulae with squares and square rootsE X AM P L E


a mx2 = b b x2 + h = k c ax2 + b = c

dffiffiffix

p= u e

ffiffiffiffiffiffiffiffiffiffix2 e

p= t f

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 + A

p= B

........................................................................................................a mx2 = b b x2 + h = k c ax2 + b = c

x2 =bm

x2 = k 2 h ax2 = c 2 b

x = 6

ffiffiffiffiffibm

rx = 6

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(k 2 h)

px2 =

c 2 ba

x = 6

ffiffiffiffiffiffiffiffiffiffiffic� ba

r

dffiffiffix

p= u e

ffiffiffiffiffiffiffiffiffiffix2 e

p= t f

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffix2 + A

p= B

x = u2 (square both sides) x 2 e = t2 x2 + A = B2

x = t2 + e x2 = B22A

x = 6ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(B2 � A)

p

Exercise 4��M /��HMake x the subject.

1 cx2 = h 2 bx2 = f 3 x2t = m 4 x2y = (a + b)

5 mx2 = (t + a) 6 x2 � a = b 7 x2 + c = t 8 x2 + y = z

9 x2 � a2 = b2 10 x2 + t 2 = m2 11 x2 + n2 = a2 12 ax2 = c

13 hx2 = n 14 cx2 = z + k 15 ax2 + b = c 16 dx2 � e = h

17 gx2 � n = m 18 x2m + y = z 19 a + mx2 = f 20 a2 + x2 = b2

Make x the subject.

21ffiffiffix

p= 2z 22

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(x� 2)

p= 3y 23

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(x + C)

p= D 24

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(ax + b)

p= c

25 b =ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(gx� t)

p26

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(d � x)

p= t 27 c =

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(n� x)

p28 g =

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi(c� x)

p

29ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(Ax + B)

p=

ffiffiffiffiD

p30 x2 = g 31 x2 = B 32 x2 � A = M

Make k the subject.

33 C � k2 = m 34 mk2 = n 35 kza

= t 36 n = a� k 2

37ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(k2 � A)

p= B 38 t =

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(m + k2)

p39 A

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(k + B)

p= M 40

ffiffiffiffiffiffiffiffiffiffiffiffiffiffi�Nk

�s= B

41ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(a2 � k2)

p= t 42 2p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(k + t)

p= 4 43

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(ak2 � b)

p= C 44 k2 + b = x2

250 Algebra 2

5.1.5 Formulae with x on both sidesE X AM P L E

Make x the subject of each formula.

a Ax� B = Cx + D b x + a =x + bc

........................................................................................................a Ax� B = Cx + D b x + a =

x + bc

Ax� Cx = D + B (x-terms on one side)c(x + a) = x + b

x(A� C) = D + B (factorise)cx + ca = x + b

x =D + BA� C

cx� x = b� ca (x-terms on one side)

x(c� 1) = b� ca (factorise)

x =b� cac� 1

Exercise 5��M /��HMake y the subject.

1 5(y� p) = 2(y + x) 2 x(y� 3) = p(3� y) 3 Ny + B = D�Ny

4 My�D = E � 2My 5 ay + b = 3b + by 6 my� c = e� ny

7 xy + 4 = 7� ky 8 Ry + D = Ty + C 9 ay� x = z + by

10 m(y + a) = n(y + b) 11 x(y� b) = y + d 12a� y

a + y= b

131� y

1 + y=

cd

14M � y

M + y=

ab

15 m( y + n) = n(n 2 y)

16 y + m =2y� 5

m17 y� n =

y + 2

n18 y + b =

ay + e

b

19ay + x

x= 4� y 20 c� dy = e� ay 21 y(a� c) = by + d

22 y(m + n) = a(y + b) 23 t � ay = s� by 24y + x

y� x= 3

25v� y

v + y=

12

26 y(b� a) = a(y + b + c) 27

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�y + x

y� x

�s= 2

28

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�z + y

z� y

�s=

13

29

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi�m(y + n)

y

�s= p 30 n� y =

4y� n

m

Exercise 6��M /��H1 A formula for calculating velocity is v = u + at.

a Rearrange the formula to express a in terms of v, u and t.

b Calculate a when v = 20, u = 4, t = 8.

2 The area of a sector of a circle is given

by the formula A =xpr2

360.

xr

Express x in terms of A, p and r.

Changing the subject of a formula 251

3 a Express k in terms of P , m and y, when P =mky.

b Express y in terms of P , m and k.

4 A formula for calculating repair bills, R, is R =n� dp

.

a Express n in terms of R, p and d.

b Calculate n when R = 400, p = 3 and d = 55.

5 The formula for the area of a circle in A = pr2.Express r in terms of A and p.

6 The volume of a cylinder is given by V = pr2h.Express h in terms of V, p and r.

7 The surface area, A, and volume, V, of a sphere are given by the

formulaeA= 4pr2 andV=43pr3.Make r the subject of each formula.

Exercise 7��M /��HMake the letter in brackets the subject.

1 ax� d = h [x] 2 zy + k = m [ y] 3 d( y + e) = f [ y]

4 m(a + k) = d [k] 5 a + bm = c [m] 6 ae2 = b [e]

7 yt 2 = z [ t ] 8 x2 � c = e [x] 9 my� n = b [ y]

10 a(z + a) = b [z] 11 ax

= d [x] 12 km

= t [k]

13 um

= n [u] 14y

x= d [x] 15 a

m= t [m]

16 dg

= n [g ] 17 tk

= (a + b) [ t ] 18 y =ve

[e]

19 c =my

[ y] 20 a2

m= b [a] 21 g(m + a) = b [m]

22 h(h + g) = x2 [g ] 23 y� t = z [ t ] 24 me2 = c [e]

25 a(y + x) = t [x] 26 uv� t 2 = y2 [v] 27 k2 + t = c [k]

28 k� w = m [w] 29 b� an = c [n] 30 m(a + y) = c [ y]

31 pq� x = ab [x] 32 a2 � bk = t [k] 33 v2z = w [z]

34 c = t � u [u] 35 xc + t = 2t [c] 36 m(n + w) = k [w]

37 v�mx = t [m] 38 c = a( y + b) [ y] 39 m(a� c) = e [c]

40 ba2 = c [a] 41 ap

= q [ p] 42 an2

= e [n]

43 hf 2

= m [ f ] 44 vx2

= n [x] 45 v� ac = t 3 [c]

46 a(a2 + y) = b3 [ y] 47 ah2 � d = b [h] 48 h(h + k) = bc [k]

49 u2 � n2 = v2 [n] 50 m(b� z) = b3 [z]

252 Algebra 2

5.2 Inequalities and regions

5.2.1 Inequality symbols

There are four inequality symbols.

l x < 4 means ‘x is less than 4’

l y > 7 means ‘y is greater than 7’

l z < 10 means ‘z is less than or equal to 10’

l t > �3 means ‘t is greater than or equal to �3’

When there are two symbols in one statement look at each part

separately.

For example, if n is an integer and 3 < n < 7,

n has to be greater than 3 but at the same time it has to be less than

or equal to 7.

So n could be 4, 5, 6 or 7 only.

E X A M P L E

Illustrate on a number line the range of values of x stated.

a x > 1

1The circle at the left-hand end

of the range is open.

This means that 1 is not included.

b x < �2

−2

The circle at �2 is filled in to

indicate that �2 is included.

c 1 < x < 4

1 4

Exercise 8��M1 Write each statement with either > or < in the box.

a 3 7 b 0 �2

c 3·1 3·01 d �3 �5

e 100mm 1m f 1 kg 1 lb

Inequalities and regions 253

2 Write the inequality displayed. Use x for the variable.

a2

b5

c100

d−2 2

e−6

f3 8

g4 7

h−5 0

i−1 3

3 Draw a number line to display these inequalities.

a x > 7 b x < 2·5 c 1 < x < 7

d 0 < x < 4 e �1 < x < 5

4 Write an inequality for each statement.

a You must be at least 16 to get married. [Use A for age.]

b Vitamin J1 is not recommended for people over 70 or for

children 3 years or under. [Use A for age recommended.]

c To cook beef the oven temperature should be between 150 �Cand 175 �C. [Use T for temperature.]

d Applicants for training as paratroopers must be at least 1·75m

tall. [Use h for height.]

5 Answer ‘true’ or ‘false’:

a n is an integer and 1 < n < 4, so n can be 2, 3 or 4.

b x is an integer and 2 < x < 5, so x can be 2, 3 or 4.

c p is an integer and p > 10, so p can be 10, 11, 12, 13 . . .

6 Which of the numbers x, below, satisfy x2 < 90?

7, 26, 10, 8·5,ffiffiffiffiffiffi95

p

7 Write one inequality to show the values of x which satisfy both of

these inequalities.

x < 7 x . 2

8 Write one inequality to show the values of x which satisfy all three

of these inequalities.

x , 5 0 , x , 6 3 < x , 10

5.2.2 Solving inequalities

Follow the same procedure that you use for solving equations

except that when you multiply or divide by a negative number you

reverse the inequality.

For example, 4 > �2 but if you multiply by �2 then �8 < 4

254 Algebra 2

It is best to avoid dividing by a negative number as in the following

example, part b.

E X A M P L E

Solve these inequalities.

a x + 11 < 4 b 8 > 13 2 x

c 2x 2 1 > 5 d x + 1 < 2x < x + 3........................................................................................................a x + 11 < 4 b 8 > 13 2 x

x < 27 (subtract 11) 8 + x > 13 (add x)

x > 5 (subtract 8)

c 2x� 1 > 5 d x + 1 < 2x < x + 3

2x > 5 + 1 (add 1) Solve the two inequalities separately.

x > 62

(divide by 2)x + 1 < 2x 2x < x + 3

x > 3

1 < x x < 3

The solution is 1 < x < 3.

Exercise 9��MSolve these inequalities.

1 x� 3 > 10 2 x + 1 < 0 3 5 > x� 7

4 2x + 1<6 5 3x� 4 > 5 6 10 < 2x� 6

7 5x < x + 1 8 2x > x� 3 9 4 + x < �4

10 3x + 1 < 2x + 5 11 2(x + 1) > x� 7 12 7 < 15� x

13 9 > 12� x 14 4� 2x < 2 15 3(x� 1)< 2(1� x)

16 7� 3x < 0 17 x35�1 18 2x

5> 3

19 2x > 0 20 x4< 0

21 The height of this picture has to be

greater than the width.

Find the range of possible values of x.

width(x + 7)

height2(x + 1)

In questions 22 to27, solve the twoinequalitiesseparately.

22 10 < 2x < x + 9 23 x < 3x + 2 < 2x + 6

24 10 < 2x� 1 < x + 5 25 3 < 3x� 1 < 2x + 7

26 x� 10 < 2(x� 1) < x 27 4x + 1 < 8x < 3(x + 2)


28 Sumitra said ‘I think of an integer.

I subtract 14.

I multiply the result by 5.

I divide by 2.

The answer is greater than the number I

thought of.’

Write an inequality and solve it to find the smallest number

Sumitra could have thought of.

Take care when there are squares and square roots in inequalities.

The equation x2 = 4 has solutions x = 62, which is correct.

For the inequality x2 < 4, you might wrongly write x < 62.Consider x = 23, say.

�3 is less than �2 and is also less than +2.

But (23)2 is not less than 4 and so

x = �3 does not satisfy the inequality x2 < 4:

The correct solution for x2 < 4 is �2 < x < 2.(23)2 = 9

E X A M P L E

a Solve the inequality

2x2 � 1 > 17.

b List the solutions which satisfy

2 < n < 14; n is a prime number........................................................................................................a 2x2 � 1 > 17

2x2 > 18

x2 > 9

x > 3 or x < �3 (Avoid the temptation

to write x > 63!)

b The prime numbers in the range

specified are 2, 3, 5, 7, 11, 13.

Exercise 10��M1 The area of the rectangle must be greater than the area of the

triangle.

Find the range of possible values of x.

x − 2

3

x + 1

4

For questions 2 to 8, list the solutions which satisfy the given condition.

2 3a + 1 < 20; a is a positive integer.

256 Algebra 2

3 b� 1 > 6; b is a prime number less than 20.

4 1 < z < 50; z is a square number.

5 2x > �10; x is a negative integer.

6 x + 1 < 2x < x + 13; x is an integer.

7 0 < 2z� 3 < z + 8; z is a prime number.

8 a2

+ 10 . a; a is a positive even number.

9 Given that 4x> 1 andx3< 1 1

3, list the possible integer values of x.

10 State the smallest integer n for which 4n > 19:

11 Given that 24 < a < 3 and 25 < b < 4, find

a the largest possible value of a2

b the smallest possible value of ab

c the largest possible value of ab

d the value of b if b2 = 25.

12 For any shape of triangle ABC, complete the statement

AB + BC AC, by writing < ,> or = inside the box.

A

B

C

13 Find a simple fraction r such that13< r <

23.

14 Find the largest prime number p such that p2 < 400:

15 Find the integer n such that n <ffiffiffiffiffiffiffiffi300

p< n + 1.

16 If f(x) = 2x 2 1 and g(x) = 102 x for what values of x is

f(x) > g(x)?

17 a The solution of x2 < 9 is 23 < x < 3.

[The square roots of 9 are 23 and 3.]

b Copy and complete.

i If x2 < 100, then < x <

ii If x2 < 81, then < x <

iii If x2 > 36, then x > or x <

Solve the inequalities.

18 x2 < 25 19 x2 < 16 20 x2 > 1

21 2x2 > 72 22 3x2 + 5 > 5 23 5x222 < 18

24 Given 2 < p < 10 and 1 < q < 4, find the range of values of

a pq bpq

c p2 q d p + q

25 If 2 r > 100, what is the smallest integer value of r ?


26 Given13

� �x<

1200

, what is the smallest integer value of x ?

27 Find the smallest integer value of x which satisfies xx > 10000:

28 What integer values of x satisfy 100 < 5x < 10000?

*29 If x is an acute angle and sin x >12, write the range of values that

x can take.

*30 If x is an acute angle and cos x >14, write the range of values that

x can take.

5.2.3 Shading regions

You can represent inequalities on a graph, particularly where two

variables (x and y) are involved.

E X A M P L E

Draw a sketch graph and shade the area which represents the set of points that

satisfy each of these inequalities.

a x . 2 b 1 < y < 5 c x + y < 8........................................................................................................

x

y

20

x =

2

x > 2

x

y

10

y = 5

1 ⩽ y ⩽ 5

5

y = 1 x

y

80

8x + y = 8

x + y ⩽ 8

In each graph, the required region is shaded.

In a, the line x = 2 is shown as a broken line to indicate that the points on the line

are not included.

In b and c points on the line are included ‘in the region’ and the lines are drawn

unbroken.

To decide which side to shade when the line is sloping, take a trial point. This

can be any point which is not actually on the line.

In c above, the trial point could be (1, 1).

Is (1, 1) in the region x + y < 8?

It satisfies x + y < 8 because 1 + 1 = 2, which is less than 8.

So below the line is x + y < 8 which is in the shaded region.

258 Algebra 2

Exercise 11��M /��HIn questions 1 to 6, describe the region which is shaded.

7 The point (1, 1), marked *, lies in the shaded region.

Use this as a trial point to describe the shaded

region as follows:

x

y

2x−

y =

3 (1, 1)

*

Is the shaded region 2x2 y . 3?

Try x = 1, y = 1. Is 2 2 1 > 3? No.

Copy and complete ‘So the shaded region is

2x 2 y 3.’

8 The point (3, 1), marked *, lies in the shaded triangle.

Use this as a trial point to write the three inequalities which

describe the shaded region.

x

y

(3,1)*

x =

8

y = −2

y = x

0

1

x

y

0

x =

3

2

x

y

0

y = 2 12

3

x

y

0

x =

6

x =

1

4

x

y

0

x =

7

y = 5 5

x

y

0

y = x

6

x

y

0

x + y = 10


9 A trial point (1, 2) lies inside the shaded triangles. Write the three

inequalities which describe each shaded region.

a

x

y

x =

0

x + y = 7

y = x

b

x

y

y = 0 x + y = 6

y = x

+ 2

For questions 10 to 27, draw a sketch graph similar to those in

question 9 and indicate the set of points which satisfy the inequalities

by shading the required region.

10 2 < x < 7 11 0 < y < 312

12 22 < x < 2 13 x < 6 and y < 4

14 0 < x < 5 and y < 3 15 1 < x < 6 and 2 < y < 8

16 23 < x < 0 and 24 < y < 2 17 y < x

18 x + y < 5 19 y > x + 2 and y < 7

20 x > 0 and y > 0 and x + y < 7 21 x > 0 and x + y < 10 and y > x

22 8 > y > 0 and x + y > 3 23 x + 2y < 10 and x > 0 and y > 0

24 3x + 2y < 18 and x > 0 and y > 0 25 x > 0, y > x22, x + y < 10

26 3x + 5y < 30 and y >x2

27 y >x2, y < 2x and x + y < 8

28 The two lines y = x + 1 and

x + y = 5 divide the graph into

four regions A, B, C, D.

Write the two inequalities which describe

each of the regions A, B, C, D.

x

y

x + y = 5

y = x

+ 1

A

B

C

D

*29 Using the same axes, draw the graphs of xy = 10 and x + y = 9

for values of x from 1 to 10.

Hence find all pairs of positive integers with products greater than

10 and sums less than 9.

260 Algebra 2

5.3 Direct and inverse proportion

5.3.1 Direct proportion

a When you buy petrol, the more you buy the more money you have

to pay. So if 2·2 litres costs 198p, then 4·4 litres will cost 396p.

The cost of petrol isdirectly proportional to the quantity bought.

To show that quantities are proportional, you use the symbol ‘}’.So in the example if the cost of petrol is c pence and the number of

litres of petrol is l, you write

c } l

The ‘}’ sign can always be replaced by ‘ = k’ where k is a

constant.

So c = kl

From above, if c = 198 when l = 2·2

then 198 = k · 2·2

k =1982·2

= 90

You can then write c = 90l, and this allows you to find the value

of c for any value of l, and vice versa.

b If a quantity z is proportional to a quantity x, you can write

z } x or z = kx

Two other expressions are sometimes used when quantities are

directly proportional. You can say

x

z

z = kx

‘z varies as x’

or ‘z varies directly as x’.

l When z and x are directly proportional the graph

connecting z and x is a straight line which passes through

the origin.

Direct and inverse proportion 261

E X A M P L E

If y varies as z, and y = 2 when z = 5, find

a the value of y when z = 6

b the value of z when y = 5.........................................................................................................Because y } z, then y = kz where k is a constant.

y = 2 when z = 5

So 2 = k · 5

k =25

So y =25z

a When z = 6, y =25

· 6 = 225.

b When y = 5, 5 =25z; z =

252

= 1212.

E X A M P L E

The value V of a diamond is proportional to

the square of its weight W.

If a diamond weighing 10 grams is worth £200, find

a the value of a diamond weighing 30 grams

b the weight of a diamond worth £5000............................................................................

V } W 2

or V = kW 2 where k is a constant.

V = 200 when W = 10

So 200 = k · 102

k = 2

So V = 2W 2

a When W = 30,

V = 2 · 302 = 2 · 900

V = £1800

So a diamond of weight 30 grams is worth £1800.

b When V = 5000,

5000 = 2 · W 2

W 2 =50002

= 2500

W =ffiffiffiffiffiffiffiffiffiffiffi2500

p= 50

So a diamond of value £5000 weighs 50 grams.

262 Algebra 2

Exercise 12��H1 Rewrite the statement connecting each pair of variables using a

constant k instead of ‘}’.Reminder: If x } d,then x = kd.

a S } e b v } t c x } z2

d y }ffiffiffix

pe T }

ffiffiffiffiL

p

2 y is proportional to t so that y = kt. If y = 6 when t = 4, calculate

the value of k and hence find

a the value of y when t = 6

b the value of t when y = 4:

3 z is proportional to m. If z = 20 when m = 4, calculate

a the value of z when m = 7

b the value of m when z = 55:

4 A varies directly as r 2. If A = 12, when r = 2, calculate

a the value of A when r = 5

b the value of r when A = 48.

5 Given that z } x, copy and complete the table.x 1 3 51

2

z 4 16

6 Given that V } r 3, copy and complete the table.r 1 2 11

2

V 4 256

7 The pressure of the water, P, at any point below the surface of the

sea varies as the depth of the point below the surface, d. If the

pressure is 200 newtons/cm2 at a depth of 3m, calculate the

pressure at a depth of 5m.

8 The distance d through which a stone falls from rest is

proportional to the square of the time taken, t. If a stone falls 45m

in 3 seconds, how far will it fall in 6 seconds? How long will it take

to fall 20m?

9 The energy, E, stored in an elastic band is proportional to the

square of the extension, x. When the elastic is extended by 3 cm,

the energy stored is 243 joules.

a What is the energy stored when the extension is 5 cm?

b What is the extension when the stored energy is 36 joules?


10 The resistance to motion of a car is proportional to the square of

the speed of the car.

a If the resistance is 4000 newtons at a speed of 20m/s, what is

the resistance at a speed of 30m/s?

b At what speed is the resistance 6250 newtons?

11 In an experiment, Julie made measurements of w and p.

w 2 5 7

p 1·6 25 68·6

Which of these laws fits the results?

p } w, p } w2, p } w3:

12 A road research organisation recently claimed that the

damage to road surfaces was proportional to the fourth

power of the axle load. The axle load of a 44-ton HGV is

about 15 times that of a car. Calculate the ratio of the

damage to road surfaces made by a 44-ton HGV and a car.

5.3.2 Inverse proportion

If you travel a distance of 200m at 10m/s, the time taken is 20 s.

If you travel the same distance at 20m/s, the time taken is 10 s.

As you double the speed, you halve the time taken.

For a fixed journey, the time taken is inversely proportional to the

speed at which you travel.

If s is inversely proportional to t, you write

s } 1t

or s = k · 1t

Sometimes youwill see ‘x variesinversely as y’. Itmeans the sameas ‘x is inverselyproportional to y’.

Notice that the product s · t is

constant.

s

t

The graph connecting s and t

is a curve.

The shape of the curve is similar

to that of y =1x.

264 Algebra 2

E X A M P L E

z is inversely proportional to t 2 and z = 4 when t = 1.

Calculate z when t = 2............................................................................z } 1

t 2or z = k · 1

t 2(k is a constant)

z = 4 when t = 1

So 4 = k

�112

�so k = 4

So z = 4 · 1t 2

When t = 2, z = 4 · 122

= 1

Exercise 13��H1 Rewrite the statements connecting the variables using a constant

of variation, k.

a x } 1y

b s } 1t 2

c t } 1ffiffiffiq

pd m varies inversely as w

e z is inversely proportional to t 2.

2 T is inversely proportional to m. If T = 12 when m = 1, findStart by writing

T =km

and then

find k.

a T when m = 2 b T when m = 24.

3 L is inversely proportional to x. If L = 24 when x = 2, find

a L when x = 8 b L when x = 32.

4 b varies inversely as e. If b = 6 when e = 2, calculate

a the value of b when e = 12

b the value of e when b = 3:

5 x is inversely proportional to y2. If x = 4 when y = 3, calculate

a the value of x when y = 1

b the value of y when x = 214:

6 p is inversely proportional toffiffiffiy

p: If p = 1·2 when y = 100,

calculate

a the value of p when y = 4

b the value of y when p = 3:


7 Given that z } 1y, copy and

complete the table.y 2 4 1

4

z 8 16

8 Given that v } 1t 2

, copy and

complete the table.t 2 5 10

v 25 14

9 e varies inversely as ( y 2 2). If e = 12 when y = 4, find

a e when y = 6 b y when e =12.

10 The volume, V, of a given mass of gas varies inversely as the

pressure, P. When V = 2m3, P = 500N=m2.

a Find the volume when the pressure is 400N=m2.

b Find the pressure when the volume is 5m3.

11 The number of hours, N, required to dig a certain hole is

inversely proportional to the number of people available, x.

When 6 people are digging, the hole takes 4 hours.

a Find the time taken when 8 people are available.

b If it takes12hour to dig the hole, how many people are there?

12 The force of attraction,F, between twomagnets varies inversely as

the square of the distance, d, between them. When the magnets

are 2 cm apart, the force of attraction is 18 newtons.How far apart

are they if the attractive force is 2 newtons?

13 The number of tiles, n, that can be pasted using one tin of tile

paste is inversely proportional to the square of the side, d, of the

tile. One tin is enough for 180 tiles of side 10 cm. How many tiles

of side 15 cm can be pasted using one tin?

14 The life expectancy, L, of a rat varies inversely as the square of the

density, d, of poison distributed around its home.

When the density of poison is 1 g=m2 the life expectancy is 50 days.

How long will the rat survive if the density of poison is

a 5 g=m2 ? b12g=m2 ?

266 Algebra 2

15 When cooking snacks in a microwave oven, a cook assumes that

the cooking time is inversely proportional to the power used. The

five levels on his microwave have the powers shown in the table.

LevelPowerused

Full 600W

Roast 400W

Simmer 200W

Defrost 100W

Warm 50W

a Escargots de Bourgogne take 5minutes on ‘Simmer’. How long

will they take on ‘Warm’?

b Escargots a la Provencale are normally cooked on ‘Roast’ for

3 minutes. How long will they take on ‘Full’?

16 Given z =kx n

, find k and n, then copy and complete the table.

x 1 2 4

z 100 1212

110

*17 Given y =kffiffiffin vp , find k and n, then copy and complete the table. ffiffi

n vp

means the nthroot of v.

v 1 4 36

y 12 6 325

5.4 Curved graphs

5.4.1 Common curves

It is helpful to know the general shape of some of the more

common curves.

The curve alsohas a line ofsymmetry.

l Quadratic curves have an x2-term as the highest

power of x.

Curved graphs 267

For example, y = 2x223x + 7 and y = 5 + 2x2 x2

a When the x2-term is positive,

the curve is -shaped.

b When the x2-term is negative

the curve is an inverted .

x

yline of symmetry

x

y

l Cubic curves have an x3-term as the highest power of x.

For example, y = x3 + 7x2 4 and y = 8x24x3

a When the x3-term is positive, the curve can be like one of the two

shown below. Notice that as x gets larger, so does y.

x

y

x

y

b When the x3-term is negative, the curve can be like one of

the two shown below. Notice that when x is large, y is large

but negative.

x

y

x

y

268 Algebra 2

l Reciprocal curves have a 1xterm.

For example, y =12x

and

y =6x

+ 5

x

y

y = 12x

The curve has a break at x = 0. The x-axis and the y-axis are called

asymptotes to the curve. The curve gets very near but never

actually touches the asymptotes.

l Exponential curves have a term involving ax, where a is

a constant.

x

y

1

y = 3x

For example, y = 3x

y =12

� �xThe x-axis is an asymptote

to the curve.

Exercise 14��M /��H1 What sort of curves are these? Give as much information as

you can.For example:‘quadratic,positive x2’

a

x

y b

x

y c

x

y

Curved graphs 269

d

x

y e

x

y f

x

y

2 Draw the general shape of these curves. (Do not draw accurate

graphs.)

a y = 3x22 7x + 11 b y = 2x c y =100x

d y = 8x2 x2 e y = 10x3 + 7x22 *f y =1x2

3 Here are the equations of the six curves in question 1, but not in the correct order.

i y =8x

ii y = 2x3 + x + 2 iii y = 5 + 3x2 x2

iv y = x226 v y = 5x vi y = 12 + 11x22x2 2 x3

Write which equations fit the curves a to f.

4 Sketch the two curves given and state the number of times the curves intersect.

a y = x3,

y = 102 x

b y = x2,

y = 102 x2

c y = x3,

y = x

*d y = 3x

y = x3

5.4.2 Plotting curved graphsE X AM P L E

Draw the graph of the function

y = 2x2 + x26, for 23 < x < 3:........................................................................................................a Make a table of x- and y-values. y

x1 2 3

468

10121416

−2−3−2−4−6

2

−1 0

y = 2x2 + x − 6

x 23 22 21 0 1 2 3

2x2 18 8 2 0 2 8 18

x 23 22 21 0 1 2 3

26 26 26 26 26 26 26 26

y 9 0 25 26 23 4 15

b Draw and label axes using suitable scales.

c Plot the points and draw a smooth curve

through them with a pencil.

d Check any points which interrupt the

smoothness of the curve.

e Label the curve with its equation.

270 Algebra 2

Common errors with graphs

Avoid these mistakes. Your curve should be smooth.

Wrong pointA series of ‘mini curves’ Flat bottom

Exercise 15��M1 a Copy and complete the table for

y = x2 + 2x.

2 a Copy and complete the table for

y = x2 2 3x.

x 23 22 21 0 1 2 3

x2 9 4 1 0 1 4 9

2x 26 24 0 4

y 3 0 0

x 23 22 21 0 1 2 3

x2 9 0 4

23x 9 0 26

y 18

b Draw the graph of y = x2 + 2x using a

scale of 2 cm for 1 unit on the x-axis

and 1 cm for 1 unit on the y-axis.

b Draw the graph of y = x2 2 3x using

the same scales as in question 1.

Draw the graphs of these functions using a scale of 2 cm for 1 unit

on the x-axis and 1 cm for 1 unit on the y-axis.

3 y = x2 + 4x, for 23 < x < 3 4 y = x2 + 2, for 23 < x < 3

5 y = x2 27, for 23 < x < 3 6 y = x2 + x22, for 23 < x < 3

7 y = x2 + 3x29, for 24 < x < 3 8 y = x2 23x24, for 22 < x < 4

9 y = x2 25x + 7, for 0 < x < 6 10 y = 2x2 26x,

for 21 < x < 5 In question 10,remember:2x2 = 2(x2).11 y = 2x2 + 3x26, for 24 < x < 2 12 y = 3x2 26x + 5,

for 21 < x < 3

13 y = 2 + x2 x2, for 23 < x < 3 14 f(x) = 123x2 x2, for 25 < x < 2

15 f(x) = 3 + 3x2 x2, for 22 < x < 5 16 f(x) = 723x22x2, for 23 < x < 3

17 f(x) = 6 + x22x2, for 23 < x < 3 18 y = 8 + 2x23x2, for 22 < x < 3

19 y = x(x24), for 21 < x < 6 20 y = (x + 1)(2x25), for 23 < x < 3:

Curved graphs 271

E X A M P L E

Draw the graph of y =12x

+ x2 6, for 1 < x < 8:

Use the graph to find approximate values for

a the minimum value of12x

+ x26

b the value of12x

+ x26, when x = 2·25:

........................................................................................................Here is the table of values.

x 1 2 3 4 5 6 8

12x

12 6 4 3 2·4 2 1·5

x 1 2 3 4 5 6 8

26 26 26 26 26 26 26 26

y 7 2 1 1 1·4 2 3·5

y

x1 2 3 4 5 6 7 80

1

2

3

4

5

6

7

0

+ x − 6y = 12x

a From the graph, the minimum value of12x

+ x2 6, (that is y)

is approximately 0·9.

b At x = 2·25, y is approximately 1·6.

Exercise 16��M /��HDraw these curves. The scales given are for one unit of x and y.

1 y =12x

, for 1 < x < 10: (Scales: 1 cm for x and y)

2 y =9x, for 1 < x < 10: (Scales: 1 cm for x and y)

3 y =12

x + 1, for 0 < x < 8: (Scales: 2 cm for x, 1 cm for y)

4 y =8

x2 4, for 24 < x < 3·5: (Scales: 2 cm for x, 1 cm for y)

5 y =x

x + 4, for 23·5 < x < 4: (Scales: 2 cm for x and y)

6 y =x + 8x + 1

, for 0 < x < 8: (Scales: 2 cm for x and y)

272 Algebra 2

7 y =10x

+ x, for 1 < x < 7: (Scales: 2 cm for x, 1 cm for y)

8 y = 3x, for 23 < x < 3. (Scales: 2 cm for x,12cm for y)

9 y =12

� �x, for 24 < x < 4. (Scales: 2 cm for x, 1 cm for y)

10 y = 5 + 3x2 x2, for 22 < x < 5: (Scales: 2 cm for x, 1 cm for y)

Find

a the maximum value of the function 5 + 3x2 x2

b the two values of x for which y = 2:

11 y =15x

+ x27, for 1 < x < 7: (Scales: 2 cm for x and y)

From your graph find

a the minimum value of y

b the y-value when x = 5·5:

12 y = x3 22x2, for 0 < x < 4: (Scales: 2 cm for x, 12cm for y)


a the y-value at x = 2·5

b the x-value at y = 15:

13 y =110

(x3 + 2x + 20), for 23 < x < 3: (Scales: 2 cm for x and y)


a the x-value where x3 + 2x + 20 = 0

b the x-value where y = 3:

*14 Draw the graph of

y =x

x2 + 1, for 26 < x < 6: (Scales: 1 cm for x, 10 cm for y)

*15 Draw the graph of

E =5000x

+ 3x for 10 < x < 80:

(Scales: 1 cm to 5 units for x and 1 cm to 25 units for E )

From the graph find

a the minimum value of E,

b the value of x corresponding to this minimum value,

c the range of values of x for which E is less than 275.

Curved graphs 273

Exercise 17��M /��H (Mixed questions)

x

1 A rectangle has a perimeter of 14 cm and length x cm. Show that the

width of the rectangle is (72 x) cm and hence that the area, A, of

the rectangle is given by the formula, A = x(72 x).

Draw the graph, plotting x on the horizontal axis with a scale of

2 cm to 1 unit, and A on the vertical axis with a scale of 1 cm to 1

unit. Take x from 0 to 7. From the graph find

a the area of the rectangle when x = 2·25 cm

b the dimensions of the rectangle when its area is 9 cm2

c the maximum area of the rectangle

d the length and width of the rectangle corresponding to the

maximum area

e what shape of rectangle has the largest area.

2 A farmer has 60m of wire fencing which he uses

to make a rectangular pen for his sheep. He uses

a stone wall as one side of the pen so the wire is

used for only three sides of the pen.width

Wall

A

length

x

a If the width of the pen is xm, what is the length (in terms

of x)?

b What is the area, A, of the pen?

c Draw a graph with area, A, on the vertical axis and the width, x,

on the horizontal axis. Take values of x from 0 to 30.

d What dimensions should the pen have if the farmer wants to

enclose the largest possible area?

3 A ball is thrown in the air so that t seconds after it is thrown,

its height h metres above its starting point is given by the function

h = 25t25t 2. Draw the graph of the function of 0 < t < 6,

plotting t on the horizontal axis with a scale of 2 cm to 1 second, and

h on the vertical axis with a scale of 2 cm for 10 metres.

Use the graph to find

a the time when the ball is at its greatest height

b the greatest height reached by the ball

c the interval of time during which the ball is at a height of more

than 30m.

274 Algebra 2

4 Consider the equation y =1x:

When x =12, y =

1

12

= 2: When x =1

100, y =

1

1100

= 100:

As the denominator of1xgets smaller, the answer gets larger.

An ‘infinitely small’ denominator gives an ‘infinitely large’ answer. The symbol forinfinity is ‘.

You write1x! ‘ as x ! 0.

Draw the graph of y =1xfor

x = 24,23,22,21,20·5,20·25, 0·25, 0·5, 1, 2, 3, 4.

(Scales: 2 cm for x and y)

5 Draw the graph of y = x +1xfor

x = 24,23,22,21,20·5,20·25, 0·25, 0·5, 1, 2, 3, 4.

(Scales: 2 cm for x and y)

6 Draw the graph of y =2x

x, for 24 < x < 7, including

x = 20·5, x = 0·5:

(Scales: 1 cm to 1 unit for x and y)

7 At time t = 0, one bacterium is placed in a culture in a

laboratory.The number of bacteria doubles every 10 minutes.

10min 10 min10min

1 2 4 8

a Draw a graph to show the growth of the bacteria

from t = 0 to t = 120min.

Use a scale of 1 cm to 10 minutes across the page

and 1 cm to 100 units up the page.

b Use your graph to estimate the time taken to

reach 800 bacteria.

*8 Draw the graph of y =x4

4x,

for x = 21, 2 34, 2 1

2, 2 1

4, 0,

14,12,34,

1, 1·5, 2, 2·5, 3, 4, 5, 6, 7.

(Scales: 2 cm to 1 unit for x, 5 cm to 1 unit for y)

a For what values of x is the gradient of the function

zero?

b For what values of x is y = 0·5?

Curved graphs 275

5.5 Graphical solution of equations

With an accurately drawn graph you can find an approximate

solution for a wide range of equations, many of which are impossible

to solve exactly by other methods.E X A M P L E

Draw the graph of the function y = 2x2 2 x23 for 22 < x < 3:

Use the graph to find approximate solutions to these equations.

a 2x22 x23 = 6 b 2x22 x = x + 5........................................................................................................

2 3

468

1012

−2

−4

1 4 x

y

−1 0

AB

y = 2x2 − x − 3

y = x + 2

y = 6

C

−2

D

2

a To solve the equation 2x2 2 x2 3 = 6,

draw the line y = 6. At the points

of intersection (A and B), y

simultaneously equals both 6 and

(2x2 2 x23):

So you can write 2x22 x23 = 6

The solutions are the x-values of the

points A and B, that is x = 21·9 and

x = 2·4 approx.

b To solve the equation 2x2 2 x = x + 5,

rearrange the equation to obtain the

function (2x2 2 x23) on the left-hand

side. In this case, subtract 3 from both sides.

2x22 x23 = x + 523

2x22 x23 = x + 2

If you now draw the line y = x + 2, the solutions of the

equation are given by the x-values of C and D, the points of

intersection, that is, x = 21·2 and x = 2·2 approx.

E X A M P L E

Assuming that the graph of y = x2 23x + 1 has been drawn, find the equation

of the line which you should draw to solve the equation x2 24x + 3 = 0.........................................................................................................Rearrange x2 24x + 3 = 0 in order to obtain (x2 23x + 1) on the

left-hand side.Remember:‘Rearrange theequation to besolved.’

x2 24x + 3 = 0

add x x2 23x + 3 = x

subtract 2 x2 23x + 1 = x22

Therefore draw the line y = x22 to solve the equation.

276 Algebra 2

Exercise 18��H1 In the diagram, the graphs of y = x222x23, y = 22 and y = x

have been drawn.

x

y

y = −2

2345

1

1 2 4−20

−4−3

y =

x2

− 2x

− 3

y = x

−2−1

−1 3

Use the graphs to find approximate solutions to these

equations.Reminder: Onlythe x-values arerequired.a x2 22x23 = 22 b x2 22x23 = x

c x2 22x23 = 0 d x2 22x +1 = 0

2 The graphs of y = x2 22, y = 2x and y = 22 x are shown.

x

y

345

−1

1

−3

678

−3 −2 1 32

y = 2

x

y = 2 − x

y = x2 − 2−2

−1

2

Use the graphs to solve these equations.

a x2 22 = 22 x

b x2 22 = 2x

c x2 22 = 2

Graphical solution of equations 277

In questions 3 to 6, use a scale of 2 cm to 1 unit for x and 1 cm to

1 unit for y.

3 Draw the graphs of the functions y = x2 22x and y = x + 1

for 21 < x < 4: Hence find approximate solutions of the

equation x2 22x = x + 1:

4 Draw the graphs of the functions y = x2 23x + 5 and y = x + 3

for 21 < x < 5: Hence find approximate solutions of the

equation x2 23x + 5 = x + 3:

5 Draw the graphs of the functions y = 6x2 x2 and y = 2x + 1 for

0 < x < 5: Hence find approximate solutions of the equation

6x2 x2 = 2x + 1:

6 a Complete the table and then draw the graph of

y = x2 24x + 1.

x 21 0 1 2 3 4

y 6 22 22 1

b On the same axes, draw the graph of y = x23.

c Find the solutions of these equations.

i x224x + 1 = x23

ii x224x + 1 = 0 [answers to 1 dp]

In questions 7 to 9, do not draw any graphs.

7 Assuming the graph of y = x2 25x has been drawn, find the

equation of the line which you should draw to solve each of these

equations.You want to getx2 � 5x on theleft-hand side.

a x2 25x = 3 b x225x = 22

c x2 25x = x + 4 d x226x = 0

e x2 25x26 = 0

8 Assuming the graph of y = x2 + x + 1 has been drawn, find the

equation of the line which you should draw to solve each of

these equations.

a x2 + x + 1 = 6 b x2 + x + 1 = 0

c x2 + x23 = 0 d x22 x + 1 = 0

e x2 2 x23 = 0

9 Assuming the graph of y = 6x2 x2 has been drawn, find the

equation of the line which you should draw to solve each of

these equations.

a 4 + 6x2 x2 = 0 b 4x2 x2 = 0

c 2 + 5x2 x2 = 0 d x226x = 3

e x2 26x = 22

278 Algebra 2

For questions 10 to 13, use scales of 2 cm to 1 unit for x and 1 cm

to 1unit for y.

10 a Complete the table and then draw the graph of

y = x2 + 3x21.

x 25 24 23 22 21 0 1 2

y 9 3 23 23 9

b By drawing other graphs, solve these equations.

i x2 + 3x21 = 0

ii x2 + 3x = 7

iii x2 + 3x23 = x

11 Draw the graph of y = x22 2x + 2 for 22 < x < 4: By drawing

other graphs, solve these equations.

a x2 22x + 2 = 8 b x2 22x + 2 = 52 x

c x2 22x25 = 0

12 Draw the graph of y = x22 7x for 0 < x < 7: Draw suitable

straight lines to solve these equations.

a x2 27x + 9 = 0 b x2 25x + 1 = 0

13 Draw the graph of y = 2x2 + 3x 2 9 for 23 < x < 2:

Draw suitable straight lines to find approximate solutions of

these equations.

a 2x2 + 3x24 = 0 b 2x2 + 2x 2 9 = 1

14 Draw the graph of y =18x

for 1 < x < 10, using scales of 1 cm to

one unit on both axes. Use the graph to solve these equations

approximately.

a18x

= x + 2 b18x

+ x = 10 c x2 = 18

15 Here are five sketch graphs (see overleaf for the other two).

x

y

y = 10x

x

y

y = 10x

1

x

y

y = 2−x

1

Graphical solution of equations 279

x

y

y = 4 − x2

2−2

4

x

y

y = x(x − 2)(x + 2)

2−2

Use the graphs to make your own sketch graphs to find the

number of solutions of each of these equations.

a10x

= 10x b 42 x2 = 10x c x(x22) (x + 2) =10x

d 22x = 10x e 42 x2 = 22x f x(x22) (x + 2) = 0

16 Draw the graph of y =12x2 26 for 24 < x < 4, taking 2 cm to

1 unit on each axis.

a Use your graph to solve approximately the equation12x2 2 6 = 1:

b Using tables or a calculator confirm that your solutions are

approximately 6ffiffiffiffiffiffi14

pand explain why this is so.

c Use your graph to find the two square roots of 8.

17 Draw the graph of y = 3x22x3 for 22 < x < 3. Use your graph

to find the range of values of k for which the equation

3x2 2 x3 = k has three solutions.

*18 Draw the graph of y = 622x2 12x3 for x = 62, 61

12, 61, 61

2, 0.

Take 4 cm to 1 unit for x and 1 cm to 1 unit for y.

Use your graph to find approximate solutions of these equations.

a12x3 + 2x2 6 = 0 b x2 1

2x3 = 0

Confirm that two of the solutions to the equation in part b are

6ffiffiffi2

pand explain why this is so.

*19 Draw the graph of y = 2x for24 < x < 4, taking 2 cm to one unit

for x and 1 cm to one unit for y. Find approximate solutions to

these equations.

a 2x = 6 b 2x = 3x c x(2x) = 1

Find also the approximate value of 22·5.

280 Algebra 2

Test yourself.................................................................................

1 Make p the subject of the formula

5p + 2q = 6 2 q

Simplify your answer as much as possible.

2 Make x the subject of this formula.

4(2x 2 y) = 2y + 5(OCR, 2003)

3 Solve the inequality 7 + n . 13 2 2n

4 a Solve the inequality 3x 2 1 < 8

b Write down the inequality shown by the following diagram.

−4 −3 −2 −1 0 1 2 3 4x

c Write down all the integers that satisfy both inequalities

shown in parts (a) and (b).

5 Write down three inequalities which together describe the shaded

region R.

10

1

0

2

3

4

2 3 4 5 6 x

y

R

6 y is inversely proportional to x2 and y = 0·8 when x = 2.

a Find an equation connecting y and x.

b Find the value of y when x = 3.(OCR, 2003)

7 y is directly proportional to the square of x.

y = 72 when x = 2

a Express y in terms of x.

b Work out the value of y when x =12.

Test yourself 281

8 b is inversely proportional to the square of a.

When b = 6, a = 2.

Calculate the value of b when a = 6.

9 a Complete the table for y = x223x + 1

x 22 21 0 1 2 3 4

y 11 1 21 1 5

b Draw the graph of y = x223x + 1

c Use your graph to find an estimate for the minimum value of y.

d Use a graphical method to find estimates of the solutions to the

equation x2 2 3x + 1 = 2x 2 4(Edexcel, 2003)

10 a Draw the graph of y =6x

for values of x from 26 to +6. 6

6−6

−6

y

x

b On the same axes draw the graph of y = x.

c Show why you can use these graphs to solve x2 = 6.

d Use these graphs to solve x2 = 6.

(OCR, 2004)

11 The region R satisfies the inequalities

x > 2, y > 1, x + y < 6

On a copy of the grid below, draw straight lines and use shading to show

the region R.

1O

1

2

3

4

5

6

7

8

2 3 4 5 6 7 8

(Edexcel, 2008)

282 Algebra 2

12 a Draw the graph of y = 2x for values of x from 22 to 3.

−2 −1 1

5

−5

10

2 30

y

x

b Use your graph to solve 2x = 6.(OCR, 2004)

13 a Which inequality is shown shaded on the grid?

Choose the correct answer.

10

1

0

2

3

4

5

6

2 3 4 5 6 x

y

y > 2 y > 2

x > 2 x > 2

b On a copy of the grid draw lines to find the region

satisfied by the three inequalities

10

1

0

2

3

4

5

6

2 3 4 5 6 x

y

y > 2

y < x + 1

x + y < 5

Label the region with the letter R.(AQA, 2007)

14 You are given that y is inversely proportional to x,

and that y = 9 when x = 4.

a i Find an equation connecting y and x.

ii Find y when x =12

b Find y when x = y.(OCR, 2008)

Test yourself 283

Functional task 4

Mr Roe’s horsesMr Roe has a farm and needs to work out the

cost of keeping his horses. The main costs are

feed, vet’s bills and keeping the horse healthy.

He has volunteers to clean the stables and look

after the horses in return for being allowed to ride

the horses every week.

FeedThe horses are given a mixture of hay and oats. The hay is free

from the farm but a 40 kg bag of oats costs £9·50. Big horses

needs more feed than small horses as shown in the feed chart.

Horse measurementBy tradition the height of a horse is given in hands and inches.

There are four inches in one hand. The formula for converting

hands and inches into centimetres is:

c =5ð4h + iÞ

2

Where c = number of centimetres

h = number of hands

i = number of inches

Mr Roe’s horses

Number Height(hands, inches)

Age(years)

1 17 hands 9

2 14 hands 2 inches 7

3 15 hands 13



6 14 hands 9

7 15 hands 15


Feed chart

Height of horse(cm)

Feed per day(kg)

135 4·2

140 4·5

145 4·8

150 5·2

155 5·6

160 6·0

165 6·5

170 7·1

284 Algebra 2

Task 1Work out the height of each horse in centimetres.

Task 2Work out the total cost for the eight horses, including

feed, vet’s bills and keeping the horse healthy.

Keeping the horse healthy

A farrier is needed to care for the horses’s feet as the

grow all the time (like human fingernails). A farrier is

needed eight times a year and it costs £20 per horse per

visit. Each horse also needs to be wormed seven times a

year at a cost of £15 per horse per visit.

Vet’s bills

Each year horses need injections against Influenza and

Tetanus. Illness or injury can be very expensive so

Mr Roe takes out insurance to cover the cost of fees.

This cost is £30 per month per horse.

Functional task 285

Sample chapter from GCSE Mathematics: Revision and Practice New 2010 Edition Higher Student Book, by...

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